Let $G$ be a simple algebraic group over $\mathbb{C}$ and $ T \subset G$ the Cartan sub-Lie-group (ie the maximal torus) and $\text{Hom}(T, \mathbb{C}^*)$ it's dual space. The lattice of weights $\Lambda_W \subset \text{Hom}(T, \mathbb{C}^*)$ is the set of $\beta \in \text{Hom}(T, \mathbb{C}^*)$ such that $\beta(H_{\alpha}) \in \mathbb{Z}$ for all co-roots $H_{\alpha}$ of the simple roots $\alpha$.
There is a known way to relate highest weights $\lambda \in \Lambda_W$ of $G$ living in a distinguished/fixed Weyl chamber (which in turn correspond to isomorphism classes of irreducible representations $V= \Gamma_{\lambda}$) to parabolic subgroups $P_{\lambda}$. The explicit way to do it can be found, eg in Fulton & Harris' Representation Theory (p. 388) as follows:
The irreducible representation $V= \Gamma_{\lambda}$ of $G$
induces an action of $G$ on the projective space
$\mathbb{P}V$, the projectivisation of $V$. Let
$p \in \mathbb{P}V$ be the point corresponding to the
eigenspace with eigenvalue $\lambda$; that's welldefined
since the eigenspace is one-dimensional.
The point $p$ is fixed under the Borel subgroup $B$
(which is iniquely determined after fixing the
positive roots in the corresponding Lie-algebra
representation), and the associated parapolic space
$P_{\lambda}$ is the the stabilizer
of $p$; the orbit $G/P_{\lambda} = G \cdot p$ is
compact and hence closed.
Simple question: Why under this construction the orbit $G \cdot p$ is not always the complete space $\mathbb{P}V$? Following considerations lead me to the conclusion that the orbit should be the whole space $\mathbb{P}V$ or equivalently that $G$ act transitively on $\mathbb{P}V$. What I'm doing wrong in my considerations below?
Let $V= \bigoplus_{\alpha \in A_{\lambda}} V_{\alpha}$ the weight decomposition of $V$ with respect the weights $\alpha: T \to \mathbb{C}^*$ of the representation $V$ for Cartan group $\text{exp}(\mathfrak{t})=T \subset G$. Since $V$ is associated with highest weight $\lambda$, every $\alpha$ with $V_{\alpha} \neq 0$ equals $\lambda$ modulo the space of roots $\Lambda_{Ad}$, ie the weights of the adjoint representation $\text{Ad}: G \to \mathfrak{gl}(\mathfrak{g})$.
These decompose $\mathfrak{g}$ into $\bigoplus_{\mathfrak{r} \in \Lambda_{Ad}} \mathfrak{g}_{\mathfrak{r}}$.
Therefore if $\alpha= \lambda + \mathfrak{r}$, $\mathfrak{r}$ root $G$ with eigenspace $\mathfrak{g}_{\mathfrak{r}} \subset \mathfrak{g}$, then for any nonzero $v \in \mathfrak{g}_{\mathfrak{r}}$, we have $V_{\alpha} = \text{exp}(v)(V_{\lambda})$ and therefore $G \cdot V_{\lambda}$ hits every direct summand $V_{\alpha}$ in the weight decomposition $V= \bigoplus_{\alpha \in A_{\lambda}} V_{\alpha}$. Therefore it hits every one-dimensional subspace in $V$, since the representation is linear. But then the action by $G$ would be transitive on $\mathbb{P}V$ and the orbit $G \cdot p$ would be the complete space $\mathbb{P}V$, or not?
But in the book is stated that in general $G/P_{\lambda} = G \cdot p$ is only a closed subspace of $\mathbb{P}V$. What is my error in the considerations above?
It's indeed a general fact that under the setting in the question, $G$ in general not acts transitively on $\mathbb{P}V$, where $V= V_{\lambda}$ is the irreducible rep of $G$ with maximal weight $\lambda$. Here Moishe Kohan's counterexample:
Consider the adjoint rep of $SL_3(\mathbb{C})$. Since the Lie-algebra $\mathfrak{sl}_3$ is $3$-dimensional and all irreps of $SL_3(\mathbb{C})$ are parametrized by dimension, we have that the irrep $V:=\mathfrak{sl}_3$ decomposes in $V= V_{-2} \oplus V_0 \oplus V_2$ with highest weight $2$. Let $p := [v_2] \in \mathbb{P}V \cong \mathbb{P}^2$ where $v_2$ eigenvector of generating weight $2$.
Then $B= \text{Stab}(p)$ and $G/B \cong \mathbb{P}^1$, so the action of $G$ on $\mathbb{P}V$ cannot be transitive.
Potential source for confusion: Let $V$ a vector space with decomposition $V= \bigoplus_{a \in A} V_a$ in one-dimensional vector spaces $V_a$.
Let $G$ a group acting linearly (ie there is a group homom $G \to GL(V)$) on $V$ with additional property that it acts transitively on the $V_a$'s: i.e. for any pair of $a, b \in A$ there exist a $g \in G$ with $g \cdot V_a =V_b$, then it not implies that $G \cdot V_a=V$. Morally, that's because the $G \cdot V_a$ is not a subvector space on $V$.
In contrast, if there ie a Lie-algebra $\mathfrak{g}$ which acts linearly on $V$ (ie there is a algebra homom $\mathfrak{g} \to End(V)$) and as before $\mathfrak{g}$ acts transitively on the eigenspaces $V_a$, this implies in contrast to the case of Lie group action that indeed $V= \mathfrak{g} \cdot V_a$.
Important remark pointed out by Callum: The 'Potential source for confusion'-part is not exactly what happens in the setting above, since neither $\mathfrak{g}$ nor $G$ really act on the "set" of eigenspaces $ \{\{V_a \}, a\in I \} $, but it is indirectly related to this in the way
that even if all weight spaces $V_{\beta}$ are contained in the same $G$-orbit - where we consider the $G$-action on the set of all lines of $V$ - this doesn't imply that the $G$-orbit of a picked $V_{\alpha}$ is the whole $V$, but on the other hand the orbit $\mathfrak{g} \cdot V_{\alpha}$ is the whole $V$ if the difference between any two weights is a root.